C ++在调用时指定函数的非模板版本

Question

I wonder, is there a way to force calling non-template function, like:

template <class T>
void foo(T&);

void foo(const int&);

void bar()
{
   int a;
   foo(a); // templated version is called, not a usual function
}

Answer 1

You may do

foo(const_cast<const int&>(a));

or

foo(static_cast<const int&>(a));

or via intermediate variable

const int& crefa = a;
foo(crefa);

or with wrapper:

foo(std::cref(a));

or alternatively specify foo :

static_cast<void(&)(const int&)>(foo)(a);

Answer 2

您只需要制作一个这样的强制转换常量：

foo(const_cast<const int &>(a));

Answer 3

I guess the problem is you have use const in the usual function. And in the template it is not const T& as parameter. That is why template version is called. Also you can use change the parameter to (const int&)a instead of simply passing a

Answer 4

foo((const int&)a);

调用int版本。